Let $\Omega$ be a domain in $\mathbb R^N$ and $u \in C(\Omega)$. If $\omega_N$ denotes the volume of $B_1(0) \subset \mathbb R^N$ and $0 < r < d(x, \partial \Omega)$, the averages are
$$
A_b(x) = \frac{1}{\omega_N r^N} \int_{B_r(x)} u(y) \ dy
$$
in the ball, and
$$
A_s(x) = \frac{1}{N \omega_N r^{N – 1}} \int_{\partial B_r(x)} u(y) \ d\sigma(y)
$$
in the sphere.
We know that if $u$ is harmonic, then $A_b(x) = A_s(x) = u(x)$.
In general, can one say something like $A_s \leq (\geq) A_b$?
I kindly ask for a hint, in case the answer is positive, or for examples, in case the answer is negative.
EDIT:
Ted's hint made me think of the following: take $u$ to be negative inside the ball and $0$ on the boundary. Then of course $A_b < 0 = A_s$. On the other hand, if $u$ is positive, it is clear that $A_b > 0 = A_s$. We conclude that the answer to the question is negative in general.
This seems to be very simple, am I missing something?
Best Answer
The answer seems to be no in the case $N=3$. Fix $\epsilon > 0$ small, and take $u$ to be a continuous function with $u(x,y,z) = 1$ when $x > \epsilon$ and $u(x,y,z) = 0$ when $x < 0$. Consider averages over $B = B_2(1,0,0)$ and $B' = B_2(-1,0,0)$ and their boundaries.
The sphere $\partial B$ has, modulo $\epsilon$, $3/4$ of a diameter in the region where $u= 1$, and by a result of Archimedes also has $3/4$ of its area there, so that average is $3/4$. But $B$ itself has more than $3/4$ of its volume in the region where $u=1$, so $$ A_b > A_s $$ for this case.
The sphere $\partial B'$ has, modulo $\epsilon$, $1/4$ of its area in the region where $u = 1$, so that average is $1/4$. But $B'$ has less than $1/4$ of its volume in that region, so $$ A_b < A_s $$ for this case.
I don't know what happens in other dimensions.